Chapter 13 - Section 13.3 - Solving Nonlinear Systems of Equations - Exercise Set - Page 948: 27

No solution

$x^2+3y^2=6$ $x^2-3y^2=10$ $x^2+3y^2+x^2-3y^2=6+10$ $2x^2=16$ $2x^2/2=16/2$ $x^2=8$ $\sqrt {x^2}=\sqrt 8$ $x = ±sqrt 8$ $x^2-3y^2=10$ $8-3y^2=10$ $8-3y^2+3y^2-10=10+3y^2-10$ $-2=3y^2$ $-2/3=3y^2/3$ $-2/3 =y^2$ $\sqrt {-2/3} =\sqrt {y^2}$ We are wanting the square root of a negative number, so there is no solution for this system.